How to Convert Ecliptical coordinates to Equatorial And Vice Versa?


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How to Convert Ecliptical coordinates to Equatorial And Vice Versa?

Feb. 25, 2023, 2:32 p.m. Mark Rusborn 2 min. to read


This article will derive equations for converting a planet's ecliptic coordinates to equatorial coordinates. This conversion is necessary for calculating primary directions - the ancient technique of event prediction.

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Notations

First, let's define terms. We will denote

Coordinate rotation

Let's denote the vector (a pointer) to the observed planet with the letter $\vec{v}$. Then the spherical coordinates of this vector in the ecliptic coordinate system will be

$$ \vec{v} = (\lambda, \delta) $$

The Cartesian coordinates of the same vector, according to equation (2) will be equal to

$$ \begin{aligned} \vec{v} &= \cos\delta\cos\lambda~\vec{X_{ecl}} \\ & + \cos\delta\sin\lambda~\vec{Y_{ecl}} \\ & + \sin\delta~\vec{Z_{ecl}} \end{aligned} $$

Here we set $R$ (the radius of the celestial sphere) to be equal to 1 for simplicity. The $X_{ecl}$-axis directs to 0° Aries, the $Y_{elc}$-axis to 0° Cancer, and the $Z_{ecl}$-axis to the northern celestial hemisphere.

The equatorial plane is inclined at an angle $-\epsilon$ relative to the plane of the ecliptic in the $Y_{ecl}Z_{ecl}$ plane.

ecliptic plane
Fig. 1 - Ecliptic and equatorial planes.

We can use the rotation matrix, which we introduced earlier, substituting ($-\sin\epsilon$) instead of $\sin(-\epsilon)$.

$$\mathbf{A}_{YZ} = \left[\begin{array} {rrr} 1 & 0 & 0 \\ 0 & \cos\epsilon & -\sin\epsilon \\ 0 & \sin\epsilon & \cos\epsilon \end{array}\right] $$

It gives us cartesian coordinates of the vector $\vec{v}$ in an equatorial coordinate system:

$$ \begin{cases} x_{eq} = \cos\delta\cos\lambda \\ y_{eq} = \cos\epsilon\cos\delta\sin\lambda - \sin\epsilon\sin\delta \\ z_{eq} = \sin\epsilon\cos\delta\sin\lambda + \cos\epsilon\sin\delta \end{cases} $$

From the conversion equation (1) it follows that

$$ \begin{cases} \tan(RA) = y_{eq} / x_{eq} \\ \sin(D) = z_{eq} \end{cases} $$

It gives us the final equations for converting $(\lambda, \delta)\rightarrow(RA, D)$:

$$ \begin{aligned} \tan(RA) = \frac{\cos\epsilon\sin\lambda - \sin\epsilon\tan\delta} {\cos\lambda} \\\ \sin(D) = \sin\epsilon\cos\delta\sin\lambda + \cos\epsilon\sin\delta \end{aligned}\tag{1} $$

Invert Conversion

For insvert conversion we change $\epsilon$ to $(-\epsilon)$. It gives us the following:

$$ \begin{aligned} \tan(\lambda) = \frac{\cos\epsilon\sin(RA) + \sin\epsilon\tan(D)} {\cos RA} \\\ \sin(\delta) = -\sin\epsilon\cos(D) \sin(RA) + \cos\epsilon\sin(D) \end{aligned}\tag{2} $$

Mark Rusborn

Mark Rusborn

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